Problem: Find $\lim_{x\to -3}\dfrac{x+3}{4-\sqrt{2x+22}}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-3$ (Choice B) B $-4$ (Choice C) C $-\dfrac{3}{4}$ (Choice D) D The limit doesn't exist
Substituting $x=-3$ into $\dfrac{x+3}{4-\sqrt{2x+22}}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since we have a rational expression with a square root on our hands, let's try to re-write it using the method of rationalization. $\begin{aligned} &\phantom{=}\dfrac{x+3}{4-\sqrt{2x+22}} \\\\ &=\dfrac{x+3}{4-\sqrt{2x+22}}\cdot\dfrac{4+\sqrt{2x+22}}{4+\sqrt{2x+22}} \gray{\text{Rationalize the denominator}} \\\\ &=\dfrac{(x+3)(4+\sqrt{2x+22})}{4^2-(2x+22)} \\\\ &=\dfrac{\cancel{(x+3)}(4+\sqrt{2x+22})}{-2\cancel{(x+3)}} \gray{\text{Cancel out common factors}} \\\\ &=\dfrac{4+\sqrt{2x+22}}{-2} \text{, for }x\neq -3 \end{aligned}$ This means that the two expressions have the same value for all $x$ -values (in their domains) except for $-3$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{x+3}{4-\sqrt{2x+22}}=\dfrac{4+\sqrt{2x+22}}{-2}$ for all $x$ -values in the interval $(-3.5,-2.5)$ except for $x=-3$. Therefore, $\lim_{x\to -3}\dfrac{x+3}{4-\sqrt{2x+22}}=\lim_{x\to -3}\dfrac{4+\sqrt{2x+22}}{-2}=-4$. (The last limit was found using direct substitution.) [I want to see how this looks graphically!] In conclusion, $\lim_{x\to -3}\dfrac{x+3}{4-\sqrt{2x+22}}=-4$.